To solve this problem, we need to find the points [tex]\((x, y)\)[/tex] on the graph of [tex]\( f(x) = \frac{x}{x+4} \)[/tex] where the line tangent to [tex]\( f \)[/tex] has a slope of [tex]\(\frac{1}{4}\)[/tex].
1. Find the derivative of [tex]\( f(x) \)[/tex]:
The derivative [tex]\( f'(x) \)[/tex] represents the slope of the tangent line at any point [tex]\( x \)[/tex] of the function [tex]\( f \)[/tex].
Given [tex]\( f(x) = \frac{x}{x+4} \)[/tex], we apply the quotient rule [tex]\( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \)[/tex]:
- [tex]\( u = x \)[/tex] and [tex]\( u' = 1 \)[/tex]
- [tex]\( v = x + 4 \)[/tex] and [tex]\( v' = 1 \)[/tex]
Therefore, [tex]\( f'(x) = \frac{(1)(x+4) - (x)(1)}{(x+4)^2} = \frac{x + 4 - x}{(x+4)^2} = \frac{4}{(x+4)^2} \)[/tex].
2. Set the derivative equal to the given slope:
We need the slope of the tangent line, [tex]\( f'(x) \)[/tex], to be [tex]\(\frac{1}{4}\)[/tex]:
[tex]\[
\frac{4}{(x+4)^2} = \frac{1}{4}
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
[tex]\[
4 \cdot 4 = (x + 4)^2
\][/tex]
[tex]\[
16 = (x + 4)^2
\][/tex]
Taking the square root of both sides, we get:
[tex]\[
x + 4 = \pm 4
\][/tex]
Thus,
[tex]\[
x + 4 = 4 \quad \text{or} \quad x + 4 = -4
\][/tex]
[tex]\[
x = 0 \quad \text{or} \quad x = -8
\][/tex]
4. Find the corresponding [tex]\( y \)[/tex]-values:
Substituting [tex]\( x = 0 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(0) = \frac{0}{0+4} = 0
\][/tex]
So, one point is [tex]\( (0, 0) \)[/tex].
Substituting [tex]\( x = -8 \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(-8) = \frac{-8}{-8+4} = \frac{-8}{-4} = 2
\][/tex]
So, another point is [tex]\( (-8, 2) \)[/tex].
Therefore, the points [tex]\((x, y)\)[/tex] on the graph of [tex]\( f \)[/tex] where the tangent line has a slope of [tex]\(\frac{1}{4}\)[/tex] are [tex]\((0, 0)\)[/tex] and [tex]\((-8, 2)\)[/tex].
The answer is: [tex]\((0,0)\)[/tex] and [tex]\((-8,2)\)[/tex].
